Question

The capacities of two capacitors are $C_{1}$ and $C_{2}$ and their respective potentials are $V_{1}$ and $V_{2} .$ If they are connected by a thin wire, then the loss of energy will be given by

• A. $\frac{C_{1} C_{2}\left(V_{1}+V_{2}\right)}{2\left(C_{1}+C_{2}\right)}$
• B. $\frac{C_{1} C_{2}\left(V_{1}-V_{2}\right)}{2\left(C_{1}+C_{2}\right)}$
• C. $\frac{C_{1} C_{2}\left(V_{1}-V_{2}\right)^{2}}{2\left(C_{1}+C_{2}\right)}$
• D. $\frac{\left(C_{1}+C_{2}\right)\left(V_{1}-V_{2}\right)}{C_{1} C_{2}}$

Answer 1

Answer: (C)

Initial energy, $U_{i}=\frac{1}{2} C_{1} V_{1}^{2}+\frac{1}{2} C_{2} V_{2}^{2} .$
Final energy $U_{f}=\frac{1}{2}\left(C_{1}+C_{2}\right) V^{2}\left(\right.$ where $\left.V=\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1} C_{2}}\right)$
Hence, energy loss,
$\Delta U=U_{i}-U_{f}=\frac{C_{1} C_{2}}{2\left(C_{1}+C_{2}\right)}\left(V_{1}-V_{2}\right)^{2}$